Callable as the default argument to dict.get without it being called if the key exists

Question

I am trying to provide a function as the default argument for the dictionary\'s get function, like this

def run():
   print \"RUNNING\"

test = {\'store\':1         


        

Answer 1

I suppose you want to have the callable applied only if the key does not exist.

There are several approaches to do so. One would be to use a defaultdict, which calls run() if key is missing.

from collections import defaultdict
def run():
   print "RUNNING"

test = {'store':1}
test.get('store', run())

test = defaultdict(run, store=1) # provides a value for store
test['store'] # gets 1
test['runthatstuff'] # gets None

Another, rather ugly one, one would be to only save callables in the dict which return the apropriate value.

test = {'store': lambda:1}
test.get('store', run)() # -> 1
test.get('runrun', run)() # -> None, prints "RUNNING".

If you want to have the return value depend on the missing key, you have to subclass defaultdict:

class mydefaultdict(defaultdict):
    def __missing__(self, key):
        val = self[key] = self.default_factory(key)
        return val

d = mydefaultdict(lambda k: k*k)
d[10] # yields 100

@mydefaultdict # decorators are fine
def d2(key):
    return -key
d2[5] # yields -5

And if you want not to add this value to the dict for the next call, you have a

def __missing__(self, key): return self.default_factory(key)

instead which calls the default factory every time a key: value pair was not explicitly added.